If a set is Carathéodory measurable in $\mathcal{R}(\mathbb{R}^d)$, is it Carathéodory measurable in $\mathcal{P}(\mathbb{R}^d)$?

Question

Let $\mathcal{R}(\mathbb{R}^d)$ be the collection of all "boxes" (or rectangles) in $\mathbb{R}^d$, let $\mathcal P(\mathbb R^d)$ be the power set of $\mathbb R^d$, and let $\mu^*$ be the Lebesgue outer measure. If a set $S$ is Carathéodory measurable in $\mathcal{R}(\mathbb{R}^d)$, meaning $$\mu^*(R) = \mu^*(R\cap S) + \mu^*(R\cap S^c) \ \ \ \ \forall R\in\mathcal{R}(\mathbb{R}^d),$$ is it Carathéodory measurable in $\mathcal{P}(\mathbb{R}^d)$, meaning $$\mu^*(R) = \mu^*(R\cap S) + \mu^*(R\cap S^c) \ \ \ \ \forall R\in\mathcal{P}(\mathbb{R}^d)?$$

Context: learning new characterizations is a useful tool throughout mathematics. Thus, I tend to conjecture new definitions in my personal notes and then seek to show they are correct. If I fail to (dis)prove one of them -as in the current case- I may post it here.

In case it is needed, my background on the topic involves reading (most of?) Hunter's Measure Theory and Tao's An Introduction to Measure Theory.

This is a reasonable question in its own right but it would be good to add further context (if any exists). Did it come up while you were trying to solve another problem? — FShrike, Aug 23 '24 at 23:10
@FShrike: the question is clear and has all the details for someone with some measure theory training need in my opinion no need for "context", for there is no ambiguity. — Mittens, Aug 24 '24 at 00:05
@Mittens That's not what I mean. The question is very clear and I already acknowledged it's a reasonable question. I'm just observing it's quite terse, too. Even just for the simple sake of avoiding other people's close-votes, it's a good idea to talk more — FShrike, Aug 24 '24 at 00:09
@FShrike: what's the matter with succinct postings? It is a treat to see every now and then clear and short postings that do not beat around the bush. — Mittens, Aug 24 '24 at 00:26
@Ramiro $\mu^*$ is the Lebesgue outer measure, and yes, $\mathcal{P}(\mathbb{R}^d)$ is the collection of all subsets of $\mathbb{R}^d$. — Sam, Aug 24 '24 at 13:28
@FShrike I did not know what context to add. Is what I've written OK? — Sam, Aug 24 '24 at 14:17
All I asked if whether or not this came from another problem or line of thought. I fail to see why that’s so controversial — FShrike, Aug 24 '24 at 15:41

Ramiro · Accepted Answer · 2024-08-24T22:28:13.527

The answer is YES. (Being $\mu^*$ the Lebesgue outer measure)

Proof: Let $S \subseteq \mathbb{R}^d$ such that $\mu^*(R) = \mu^*(R\cap S) + \mu^*(R\cap S^c) \ \ \ \ \forall R\in\mathcal{R}(\mathbb{R}^d)$.

For each $n \in \mathbb{N}$, $n \geqslant 1$, let $R_n = [-n, n]^d$. Then $\mu^*(R_n) = \mu^*(R_n\cap S) + \mu^*(R_n\cap S^c)$. Since $\mu(R_n) <+ \infty$ and $\mu^*$ is the outer measure induced by $\mu$, we can apply the following result

Let $\mu$ be a premeasure (or a measure) on a space $X$ such that $\mu(X) < \infty$. Then $E \subseteq X$ is Caratheodory measurable iff $ \mu^*(E)+ \mu^*(E^C) = \mu(X)$.

(the proof of this result can be found in this question)

So, we can conclude that $S\cap R_n$ is $\mu^*$-measurable in $R_n$ (that is, in $\mathcal{P}(R_n)$).

Since $R_n$ is $\mu^*$-measurable in $\mathbb{R}^d$, we have that $S\cap R_n$ is $\mu^*$-measurable in $\mathbb{R}^d$ (that is, in $\mathcal{P}(\mathbb{R}^d)$).

Since $S = \bigcup_{n=1}^\infty (S \cap R_n)$, we have that $S$ is $\mu^*$-measurable in $\mathbb{R}^d$ (that is, in $\mathcal{P}(\mathbb{R}^d)$).

If a set is Carathéodory measurable in $\mathcal{R}(\mathbb{R}^d)$, is it Carathéodory measurable in $\mathcal{P}(\mathbb{R}^d)$?

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